在keil4环境下编写的贪吃蛇在神舟四号开发板上运行,不显...
时间:10-02
整理:3721RD
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#include "stm32f10x.h"
#include "gui.h"
#include "delay.h"
#include "snake.h"
#include "lcd_ssd1289.h"
#include "stdlib.h"
#include "key.h"
int score;
struct Food
{
int x;/*食物的横坐标*/
int y;/*食物的纵坐标*/
int yes;/*判断是否要出现食物的变量*/
}food;
/*食物的结构体*/
struct Snake
{
int x[M];
int y[M];
int node;/*蛇的节数*/
int direction;/*蛇移动方向*/
int life;/* 蛇的生命,0活着,1死亡*/
}snake;
/*开始画面,左上角坐标为(10,40),右下角坐标为(310,230)的围墙*/
void DrawK(void)
{
GUI_FillRect(0,0,320,10);
GUI_FillRect(0,0,10,240);
GUI_FillRect(0,230,320,240);
GUI_FillRect(310,0,320,240);
}
/*玩游戏具体过程*/
void GamePlay(void)
{
TYPEDEF_KEY key;
unsigned char i;
key = KEY_NULL;
key = GetKey();
rand();/*随机数发生器*/
food.yes=1;/*1表示需要出现新食物,0表示已经存在食物*/
snake.life=0;/*活着*/
snake.direction=1;/*方向往右*/
snake.x[0]=100;snake.y[0]=100;/*蛇头*/
snake.x[1]=110;snake.y[1]=100;
snake.node=2;/*节数*/
PrScore();/*输出得分*/
while(1)
{
while(key == KEY_NULL)/*在没有按键的情况下,蛇自己移动身体*/
{
if(food.yes==1)/*需要出现新食物*/
{
food.x=rand()%400+60;
food.y=rand()%350+60;
while(food.x%10!=0)/*食物随机出现后必须让食物能够在整格内,这样才可以让蛇吃到*/
food.x++;
while(food.y%10!=0)
food.y++;
food.yes=0;/*画面上有食物了*/
}
if(food.yes==0)/*画面上有食物了就要显示*/
{
GUI_FillRect(food.x,food.y,food.x+10,food.y-10);
}
for(i=snake.node-1;i>0;i--)/*蛇的每个环节往前移动,也就是贪吃蛇的关键算法*/
{
snake.x[i]=snake.x[i-1];
snake.y[i]=snake.y[i-1];
}
/*1,2,3,4表示右,左,上,下四个方向,通过这个判断来移动蛇头*/
switch(snake.direction)
{
case 1:snake.x[0]+=10;break;
case 2: snake.x[0]-=10;break;
case 3: snake.y[0]-=10;break;
case 4: snake.y[0]+=10;break;
}
for(i=3;i<snake.node;i++)/*从蛇的第四节开始判断是否撞到自己了,因为蛇头为两节,第三节不可能拐过来*/
{
if(snake.x[i]==snake.x[0]&&snake.y[i]==snake.y[0])
{
GameOver();/*显示失败*/
snake.life=1;
break;
}
}
if(snake.x[0]<10||snake.x[0]>310||snake.y[0]<10||
snake.y[0]>230)/*蛇是否撞到墙壁*/
{
GameOver();/*本次游戏结束*/
snake.life=1; /*蛇死*/
}
if(snake.life==1)/*以上两种判断以后,如果蛇死就跳出内循环,重新开始*/
break;
if(snake.x[0]==food.x&&snake.y[0]==food.y)/*吃到食物以后*/
{
/*把画面上的食物东西去掉*/
GUI_ClearRect(food.x,food.y,food.x+10,food.y-10);
snake.x[snake.node]=-20;snake.y[snake.node]=-20;
/*新的一节先放在看不见的位置,下次循环就取前一节的位置*/
snake.node++;/*蛇的身体长一节*/
food.yes=1;/*画面上需要出现新的食物*/
score+=10;
PrScore();/*输出新得分*/
}
/*画出蛇*/
for(i=0;i<snake.node;i++)
{
GUI_FillRect(snake.x[i],snake.y[i],snake.x[i]+10,
snake.y[i]-10);
}
Delay_ms(1000);/*游戏速度自己调整*/
//去除去除蛇的的最后一节*
GUI_ClearRect(snake.x[snake.node-1],snake.y[snake.node-1],
snake.x[snake.node-1]+10,snake.y[snake.node-1]-10);
/*判断是否往相反的方向移动*/
if(key==KEY_1&&snake.direction!=4)
snake.direction=3;
else
if(key==KEY_2&&snake.direction!=2)
snake.direction=1;
else
if(key==KEY_3&&snake.direction!=1)
snake.direction=2;
else
if(key==KEY_4&&snake.direction!=3)
snake.direction=4;
}
}
}
/*游戏结束*/
void GameOver(void)
{
// GUI_Clear();
GUI_DispStringAt("GAME OVER!",100,80);
PrScore();/*输出得分*/
}
/*输出成绩*/
void PrScore(void)
{
GUI_DispDecAt( score, 100,20,10);
}
#include "gui.h"
#include "delay.h"
#include "snake.h"
#include "lcd_ssd1289.h"
#include "stdlib.h"
#include "key.h"
int score;
struct Food
{
int x;/*食物的横坐标*/
int y;/*食物的纵坐标*/
int yes;/*判断是否要出现食物的变量*/
}food;
/*食物的结构体*/
struct Snake
{
int x[M];
int y[M];
int node;/*蛇的节数*/
int direction;/*蛇移动方向*/
int life;/* 蛇的生命,0活着,1死亡*/
}snake;
/*开始画面,左上角坐标为(10,40),右下角坐标为(310,230)的围墙*/
void DrawK(void)
{
GUI_FillRect(0,0,320,10);
GUI_FillRect(0,0,10,240);
GUI_FillRect(0,230,320,240);
GUI_FillRect(310,0,320,240);
}
/*玩游戏具体过程*/
void GamePlay(void)
{
TYPEDEF_KEY key;
unsigned char i;
key = KEY_NULL;
key = GetKey();
rand();/*随机数发生器*/
food.yes=1;/*1表示需要出现新食物,0表示已经存在食物*/
snake.life=0;/*活着*/
snake.direction=1;/*方向往右*/
snake.x[0]=100;snake.y[0]=100;/*蛇头*/
snake.x[1]=110;snake.y[1]=100;
snake.node=2;/*节数*/
PrScore();/*输出得分*/
while(1)
{
while(key == KEY_NULL)/*在没有按键的情况下,蛇自己移动身体*/
{
if(food.yes==1)/*需要出现新食物*/
{
food.x=rand()%400+60;
food.y=rand()%350+60;
while(food.x%10!=0)/*食物随机出现后必须让食物能够在整格内,这样才可以让蛇吃到*/
food.x++;
while(food.y%10!=0)
food.y++;
food.yes=0;/*画面上有食物了*/
}
if(food.yes==0)/*画面上有食物了就要显示*/
{
GUI_FillRect(food.x,food.y,food.x+10,food.y-10);
}
for(i=snake.node-1;i>0;i--)/*蛇的每个环节往前移动,也就是贪吃蛇的关键算法*/
{
snake.x[i]=snake.x[i-1];
snake.y[i]=snake.y[i-1];
}
/*1,2,3,4表示右,左,上,下四个方向,通过这个判断来移动蛇头*/
switch(snake.direction)
{
case 1:snake.x[0]+=10;break;
case 2: snake.x[0]-=10;break;
case 3: snake.y[0]-=10;break;
case 4: snake.y[0]+=10;break;
}
for(i=3;i<snake.node;i++)/*从蛇的第四节开始判断是否撞到自己了,因为蛇头为两节,第三节不可能拐过来*/
{
if(snake.x[i]==snake.x[0]&&snake.y[i]==snake.y[0])
{
GameOver();/*显示失败*/
snake.life=1;
break;
}
}
if(snake.x[0]<10||snake.x[0]>310||snake.y[0]<10||
snake.y[0]>230)/*蛇是否撞到墙壁*/
{
GameOver();/*本次游戏结束*/
snake.life=1; /*蛇死*/
}
if(snake.life==1)/*以上两种判断以后,如果蛇死就跳出内循环,重新开始*/
break;
if(snake.x[0]==food.x&&snake.y[0]==food.y)/*吃到食物以后*/
{
/*把画面上的食物东西去掉*/
GUI_ClearRect(food.x,food.y,food.x+10,food.y-10);
snake.x[snake.node]=-20;snake.y[snake.node]=-20;
/*新的一节先放在看不见的位置,下次循环就取前一节的位置*/
snake.node++;/*蛇的身体长一节*/
food.yes=1;/*画面上需要出现新的食物*/
score+=10;
PrScore();/*输出新得分*/
}
/*画出蛇*/
for(i=0;i<snake.node;i++)
{
GUI_FillRect(snake.x[i],snake.y[i],snake.x[i]+10,
snake.y[i]-10);
}
Delay_ms(1000);/*游戏速度自己调整*/
//去除去除蛇的的最后一节*
GUI_ClearRect(snake.x[snake.node-1],snake.y[snake.node-1],
snake.x[snake.node-1]+10,snake.y[snake.node-1]-10);
/*判断是否往相反的方向移动*/
if(key==KEY_1&&snake.direction!=4)
snake.direction=3;
else
if(key==KEY_2&&snake.direction!=2)
snake.direction=1;
else
if(key==KEY_3&&snake.direction!=1)
snake.direction=2;
else
if(key==KEY_4&&snake.direction!=3)
snake.direction=4;
}
}
}
/*游戏结束*/
void GameOver(void)
{
// GUI_Clear();
GUI_DispStringAt("GAME OVER!",100,80);
PrScore();/*输出得分*/
}
/*输出成绩*/
void PrScore(void)
{
GUI_DispDecAt( score, 100,20,10);
}
不知道什么情况
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求大神编写宝马版的LPC1768的贪吃蛇游戏啊。感激涕零。